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If 9^(2x + 5) = 27^(3x - 10), then x =
[#permalink]
20 Nov 2017, 11:36

1

1

Bookmarks

Question Stats:

If \(9^{(2x + 5)} = 27^{(3x - 10)}\), then x =

A. 3

B. 6

C. 8

D. 12

E. 15

Kudos for correct solution.

A. 3

B. 6

C. 8

D. 12

E. 15

Kudos for correct solution.

If 9^(2x + 5) = 27^(3x − 10), then x =
[#permalink]
26 Nov 2017, 09:25

If 9^(2x + 5) = 27^(3x − 10), then x =

A. 3

B. 6

C. 8

D. 12

E. 15

Kudos for correct solution.

A. 3

B. 6

C. 8

D. 12

E. 15

Kudos for correct solution.

Retired Moderator

Joined: **10 Apr 2015 **

Posts: **6218**

Given Kudos: **136 **

Re: If 9^(2x + 5) = 27^(3x - 10), then x =
[#permalink]
29 Nov 2017, 10:01

1

Bunuel wrote:

If \(9^{(2x + 5)} = 27^{(3x - 10)}\), then x =

A. 3

B. 6

C. 8

D. 12

E. 15

A. 3

B. 6

C. 8

D. 12

E. 15

First, we need a COMMON BASE

Here, that common base will be 3 (since we can rewrite 9 and 27 as powers of 3)

Give: 9^(2x + 5) = 27^(3x - 10)

Rewrite bases as follows: (3²)^(2x + 5) = (3³)^(3x - 10)

Apply Power of a Power rule to get: 3^(4x + 10) = 3^(9x - 30)

So, (4x + 10) = (9x - 30)

Add 30 to both sides: 4x + 40 = 9x

Subtract 4x from both sides: 40 = 5x

Solve: x = 8

Answer: C

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Re: If 9^(2x + 5) = 27^(3x − 10), then x =
[#permalink]
30 Nov 2017, 06:16

Bunuel wrote:

If 9^(2x + 5) = 27^(3x − 10), then x =

A. 3

B. 6

C. 8

D. 12

E. 15

A. 3

B. 6

C. 8

D. 12

E. 15

The above equation can be written as \(3^{2(2x + 5)} = 3^{3(3x - 10)}\)

since the base are equal we can opt out, therefore we have

2(2x + 5) = 3(3x - 10)

4x + 10 = 9x - 30

x = 8

If 9^2X+5
[#permalink]
21 Nov 2018, 08:41

Expert Reply

1

Bookmarks

If \(9^{2x+5} = 27^{3x-10}\), then x=

A. 3

B. 6

C. 8

D. 12

E. 15

A. 3

B. 6

C. 8

D. 12

E. 15

Retired Moderator

Joined: **10 Apr 2015 **

Posts: **6218**

Given Kudos: **136 **

Re: If 9^2X+5
[#permalink]
21 Nov 2018, 10:06

1

Carcass wrote:

If \(9^{2x+5} = 27^{3x-10}\), then x=

A. 3

B. 6

C. 8

D. 12

E. 15

A. 3

B. 6

C. 8

D. 12

E. 15

Rewrite each side with the SAME BASE.

So, we can replace 9 with 3², and replace 27 with 3³ to get: \((3²)^{2x+5} = (3³)^{3x-10}\)

Apply the Power of a Power law to get: \(3^{4x+10} = 3^{9x-30}\)

This means that \(4x+10=9x-30\)

Add 30 to both sides to get: \(4x+40=9x\)

Subtract 4x from both sides to get: \(40=5x\)

Solve: x = 8

Answer: C

RELATED VIDEOS FROM OUR COURSE

Manager

Joined: **07 Aug 2016 **

Posts: **59**

Given Kudos: **0 **

Re: If 9^2X+5
[#permalink]
23 Nov 2018, 12:28

Carcass wrote:

If \(9^{2x+5} = 27^{3x-10}\), then x=

A. 3

B. 6

C. 8

D. 12

E. 15

A. 3

B. 6

C. 8

D. 12

E. 15

3^2(2x + 5) = 3^3(3x - 10)

4x + 10 = 9x - 30

-5x = -40

x = 8

Re: If 9^(2x + 5) = 27^(3x - 10), then x =
[#permalink]
11 Jul 2022, 22:24

1

9^(2x + 5) = 27^(3x − 10)

3^2(2x + 5) = 3^3 (3x - 10)

2(2x + 5) = 3(3x - 10)

4x + 10 = 9x - 30

40 = 5x

x = 8

3^2(2x + 5) = 3^3 (3x - 10)

2(2x + 5) = 3(3x - 10)

4x + 10 = 9x - 30

40 = 5x

x = 8

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Schools: **XLRI Jamshedpur, India - Class of 2014**

GMAT 1: **700 Q51 V31**

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Re: If 9^(2x + 5) = 27^(3x 10), then x =
[#permalink]
19 Sep 2022, 09:32

1

Given that \(9^{(2x + 5)}\) = \(27^{(3x − 10)}\) and we need to find the value of x

\(9^{(2x + 5)}\) = \(27^{(3x − 10)}\)

=> \((3^2)^{(2x + 5)}\) = \((3^3)^{(3x − 10)}\)

=> \(3^{2*(2x + 5)}\) = \(3^{3*(3x − 10)}\)

=> \(3^{(4x + 10)}\) = \(3^{(9x − 30)}\)

Since base is same (3) in both left hand side and right hand side so even the power will be the same

=> 4x + 10 = 9x − 30

=> 9x - 4x = 10 + 30 = 40

=> 5x = 40

=> x = \(\frac{40}{5}\) = 8

So, Answer will be C

Hope it helps!

\(9^{(2x + 5)}\) = \(27^{(3x − 10)}\)

=> \((3^2)^{(2x + 5)}\) = \((3^3)^{(3x − 10)}\)

=> \(3^{2*(2x + 5)}\) = \(3^{3*(3x − 10)}\)

=> \(3^{(4x + 10)}\) = \(3^{(9x − 30)}\)

Since base is same (3) in both left hand side and right hand side so even the power will be the same

=> 4x + 10 = 9x − 30

=> 9x - 4x = 10 + 30 = 40

=> 5x = 40

=> x = \(\frac{40}{5}\) = 8

So, Answer will be C

Hope it helps!

Re: If 9^(2x + 5) = 27^(3x - 10), then x =
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11 Jun 2024, 10:31

Hello from the GRE Prep Club BumpBot!

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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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